Have you heard of the wine-water mixing puzzle? (Spoiler alert: solution is given right after it)

Two pitchers contain equal amounts of different liquids (wine and water, for example). A glassful from the water pitcher is poured and mixed into the wine, and then a glassful from the now diluted wine is poured back into the water. Which pitcher is now more contaminated, i. e. contains more of the other liquid?

Before you whip out a pen and the back of an envelope, or throw your arms up in the air, here is the surprising answer: both are equally contaminated! One can reason algebraically (as on Wikipedia) or simply notice that whatever amount of wine is ultimately in the water must have displaced an equal amount of water into the wine, if both pitchers are back to their initial volume.

Astonishingly, this has a very nice geometric picture in terms of two straight angles "contaminating each other":

Through the rotation, some of liquid 1 (wine) lands in the half-plane of pitcher 2 and vice versa. The respective amounts are represented by angles, which by the vertical angle theorem must be equal.

The pitchers even do not have to have the same volume, as long as each ends up with its initial volume:

Don't you just wish math had more of these simple, intuitive, eye-opening picture moments? Save that wish, because it does!

(Careful though: the amounts exchanged, as represented by the opposing acute angles, are not the amounts of liquid poured, which are slightly larger. After pouring water into wine and mixing, some of that water is transferred back into its original pitcher. The intermediate stage, as well as the process of mixing, do not have an easy geometric picture as far as I can see. But what ultimately matters in the puzzle is only the final state in comparison to the beginning.)